Question

How many 4 letter words can be formed from the letter of the word ‘COMBINATION’ where all the 4 letters are different?

Answer 1

Hint: Before jumping to solve these types of problems, calculate correctly the number of different types of letters that are present in that word.
The number of repeated letters and how many times they are repeated.
Formulas used:
No of m letter words can be formed from a n letter word is = \[^{n}{{C}_{m}}\]
No arrangements in a m letter word can be possible in \[m!\] ways.

Complete step-by-step answer:
In the word ‘COMBINATION’ we are getting
No of letters: -
 C…….1 time
 O …….2 times
 M……1 times
 B…….1 times
 I……...2 times
  N…….2 times
  A…….1 times
  T…….1 times

We can form so many numbers of words using 4 letters. They are of three types.
All the letters are different.
2 letters are same of kind and 2 letters are of different kinds.
2 letters are the same and 2 are different.
Here we will consider the 1st case. i.e. all the four letters are different.
So, Number of types of different letters in the word ‘COMBINATION’ is 8.
From this, 8 different types of letters we are going to make words of 4 letters.
So, number of words we can form by this is = \[^{8}{{C}_{4}}\]
Now, those 4 letters can interchange their place in between themselves.
So, the number of ways they can interchange is = \[4!\]

As a result, the total number of ways we can make different words with 4 different letters is \[^{8}{{C}_{4}}.4!\]
 =\[\dfrac{8!}{4!.4!}.4!\]
 = \[\dfrac{8!}{4!}\]
 = \[\dfrac{8.7.6.5.4!}{4!}\]
        = 8 X 7 X 6 X 5
        = 1680
Hence, 1680 numbers of 4 letter words can be formed from the letter of the word ‘COMBINATION’ where all the 4 letters are different.

Note: For case-2 you have to check the condition of 2 same kind letters and 2 different kinds of same letters.
In that case, we have 3 repeated letters so no words. \[^{3}{{C}_{2}}\]
And, for 2 alike 2 different we have \[\dfrac{4!}{2!2!}\]
So, Total number of that kind is = \[^{3}{{C}_{2}}.\dfrac{4!}{2!2!}\]= 12
For case-3 you have to check the condition of 2 same kind letters and 2 different letters.
In that case, we will have = \[^{3}{{C}_{1}}{{.}^{5}}{{C}_{2}}.\dfrac{4!}{2!}\] = 360
So, for that all cases total words will be (1680 + 12 +360) =2052